POJ 3481 Double Queue（STL之双向优先队列）

怼烎@ 2022-06-12 13:10 159阅读 0赞

The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using modern information technologies. As usual, each client of the bank is identified by a positive integer *K* and, upon arriving to the bank for some services, he or she receives a positive integer priority *P*. One of the inventions of the young managers of the bank shocked the software engineer of the serving system. They proposed to break the tradition by sometimes calling the serving desk with the **lowest** priority instead of that with the highest priority. Thus, the system will receive the following types of request:

<table style="border-collapse:collapse"> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>The system needs to stop serving</td> 
  </tr> 
  <tr> 
   <td>1&nbsp;<em>KP</em></td> 
   <td>Add client&nbsp;<em>K</em>&nbsp;to the waiting list with priority&nbsp;<em>P</em></td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>Serve the client with the highest priority and drop him or her from the waiting list</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>Serve the client with the lowest priority and drop him or her from the waiting list</td> 
  </tr> 
 </tbody> 
</table>

Your task is to help the software engineer of the bank by writing a program to implement the requested serving policy.

Input

Each line of the input contains one of the possible requests; only the last line contains the stop-request (code 0). You may assume that when there is a request to include a new client in the list (code 1), there is no other request in the list of the same client or with the same priority. An identifier *K* is always less than 106, and a priority *P* is less than 107. The client may arrive for being served multiple times, and each time may obtain a different priority.

Output

For each request with code 2 or 3, the program has to print, in a separate line of the standard output, the identifier of the served client. If the request arrives when the waiting list is empty, then the program prints zero (0) to the output.

Sample Input

2
    1 20 14
    1 30 3
    2
    1 10 99
    3
    2
    2
    0

Sample Output

0
    20
    30
    10
    0

题解：

比赛的时候没有多想，就直接用两个优先队列怼了一次就ac了。。。比完赛一搜。。我去居然是什么平衡树，我什么时候会了平衡树hhhhhh，stl真是个神器

代码：

#include<iostream>
    #include<stdio.h>
    #include<map>
    #include<algorithm>
    #include<math.h>
    #include<queue>
    #include<stack>
    #include<string>
    #include<cstring>
    using namespace std;
    struct node
    {
        int v;
        int p;
        friend bool operator<(const node& x,const node& y)//为了建一个最大堆
        {
            return x.p<y.p;
        }
        friend bool operator>(const node& x,const node& y)//建一个最小堆
        {
            return x.p>y.p;
        }
    };
    priority_queue<node,vector<node>,less<node> >q1;
    priority_queue<node,vector<node>,greater<node> >q2;
    int p1[10000005];//看这个数是否在队列中
    int p2[1000005];//看这个优先级是否在队列中
    int main()
    {
        int i,j,n,m,x,va,pr;
        memset(p1,0,sizeof(p1));
        memset(p2,0,sizeof(p2));
        node t;
        while(scanf("%d",&x)!=EOF&&x)
        {
            if(x==1)
            {
                scanf("%d%d",&va,&pr);
                t.p=pr;
                t.v=va;
                q1.push(t);
                q2.push(t);
                p1[pr]++;
                p2[va]++;//记录信息
            }
            else if(x==2)
            {
                if(q1.empty())
                {
                    printf("0\n");
                    continue;
                }
                t=q1.top();
                while(!q1.empty()&&!p1[t.p]&&!p2[t.v])//删除之前删掉的信息
                {
                    q1.pop();
                    t=q1.top();
                }
                if(q1.empty())
                {
                    printf("0\n");
                    continue;
                }
                else
                {
                    printf("%d\n",t.v);
                    p1[t.p]--;
                    p2[t.v]--;
                    q1.pop();
                }
            }
            else
            {
                if(q2.empty())
                {
                    printf("0\n");
                    continue;
                }
                t=q2.top();
                while(!q2.empty()&&!p1[t.p]&&!p2[t.v])//删除之前删掉的信息
                {
                    q2.pop();
                    t=q2.top();
                }
                if(q2.empty())
                {
                    printf("0\n");
                    continue;
                }
                else
                {
                    printf("%d\n",t.v);
                    p1[t.p]--;
                    p2[t.v]--;
                    q2.pop();
                }
            }
        }
        return 0;
    }